Show that the set $\{x \in \mathbb{R}| f(x)=x\}$ is closed, assuming that $f:\mathbb{R} \to \mathbb{R}$ is continuous. So here is my attempt:
Let $\{x_k\} \subseteq \mathbb{R}$ such that $x_k \in \{x \in \mathbb{R}| f(x)=x\}$ and $x_k \to x$ as $k \to \infty$.Then, as $f:\mathbb{R} \to \mathbb{R}$ is continuous by a theorem in my text, the sequence $\{y_k\}$ where $\{y_k\} = f(x_k)$ converges to $f(x)$. So we have that $f(x) = \lim_{x \to \infty} f(x_k) =x $. So by definition $x$ is in the set, and we can conclude now that since the set contains all of its limit points, that the set is a closed subset of $\mathbb{R}$.
I have a feeling I made some flaws in logic somewhere around here, any help is appreciated. I only want some hints on what I did wrong so I can think about the problem longer instead of just having the solution.
Best Answer
Here it is another way to approach it for the sake of curiosity.
Consider the function $g(x) = f(x) - x$. Since $f$ is continuous, $g$ is continuous as well.
Once $\{0\}$ is closed, we conclude that $g^{-1}(\{0\}) = \{x\in\mathbb{R}\mid f(x) = x\}$ is closed, and we are done.
Hopefully this helps!